Problem: Simplify and expand the following expression: $ \dfrac{2}{2n + 10}- \dfrac{1}{4n - 20}- \dfrac{3n}{n^2 - 25} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{2}{2n + 10} = \dfrac{2}{2(n + 5)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{1}{4n - 20} = \dfrac{1}{4(n - 5)}$ We can factor the quadratic in the third term: $ \dfrac{3n}{n^2 - 25} = \dfrac{3n}{(n + 5)(n - 5)}$ Now we have: $ \dfrac{2}{2(n + 5)}- \dfrac{1}{4(n - 5)}- \dfrac{3n}{(n + 5)(n - 5)} $ The least common multiple of the denominators is: $ 8(n + 5)(n - 5)$ In order to get the first term over $8(n + 5)(n - 5)$ , multiply by $\dfrac{4(n - 5)}{4(n - 5)}$ $ \dfrac{2}{2(n + 5)} \times \dfrac{4(n - 5)}{4(n - 5)} = \dfrac{8(n - 5)}{8(n + 5)(n - 5)} $ In order to get the second term over $8(n + 5)(n - 5)$ , multiply by $\dfrac{2(n + 5)}{2(n + 5)}$ $ \dfrac{1}{4(n - 5)} \times \dfrac{2(n + 5)}{2(n + 5)} = \dfrac{2(n + 5)}{8(n + 5)(n - 5)} $ In order to get the third term over $8(n + 5)(n - 5)$ , multiply by $\dfrac{8}{8}$ $ \dfrac{3n}{(n + 5)(n - 5)} \times \dfrac{8}{8} = \dfrac{24n}{8(n + 5)(n - 5)} $ Now we have: $ \dfrac{8(n - 5)}{8(n + 5)(n - 5)} - \dfrac{2(n + 5)}{8(n + 5)(n - 5)} - \dfrac{24n}{8(n + 5)(n - 5)} $ $ = \dfrac{ 8(n - 5) - 2(n + 5) - 24n} {8(n + 5)(n - 5)} $ Expand: $ = \dfrac{8n - 40 - 2n - 10 - 24n}{8n^2 - 200} $ $ = \dfrac{-18n - 50}{8n^2 - 200}$ Simplify: $ = \dfrac{-9n - 25}{4n^2 - 100}$